Weak convergence of the image of an $L^1$ converging sequence under a convex function Suppose that $u_k$ is a sequence of $L^1$ functions defined on a compact $K\subset R^n$ and a function $f:[0, \infty)\to[0, \infty)$ with the following properties


*

*$u_k\ge 0$

*$\|u_k\|_{L^1}=\int u_k=1$

*$u_k\to u$ strongly in $L^1$

*$f$ is convex, $f(0)=0$ and has superlineair growth at $+\infty$ (that is: $\lim_{z\to+\infty} \frac{f(z)}{z}=+\infty$)

*$\int f(u_k)< C$ for all $k$


Note that the conditions on $f$ imply that the functional $v\mapsto\int f(v)$ is lower semicontinuous with respect to weak $L^1$ convergence.
Does this imply that $f(u_k)$ converges to $f(u)$ weakly in $L^1$, possibly under the extra assumption that $\int f(u_k)\to \int f(u)$?
EDIT: Thanks for the answers, both were helpful and received an upvote.
 A: I think I can even manage to prove strong convergence in $L^{1}$. I'm going to show that every subsequence has a further subsequence which converges. So, passing to subsequence, we may assume that $u_{k} \to u$ almost everywhere. Since convex functions are continuous (maybe not at $0$)*, we have convergence $f(u_{k}) \to f(u)$ almost evyrewhere. With assumption $\int f(u_{k}) \to \int f(u)$, we can use Scheffe's theorem which states that if a sequence of densities converges almost everywhere to a density, then it is, in fact, strong convergence in $L^{1}$. Of course, we only have $\int f(u_{k}) \to \int f(u)$ instead of $\int f(u_{k}) = \int f(u)$ but it requires no modification of standard proof.
I hope it's quite clear and, what is even more important, doesn't contain any mistake.
*Since $f(0)=0$ and $f$ is nonnegative and convex, then it is also continuous at $0$.
As commented by Pietro Majer, only continuity of $f$ really matters.
A: With the extra assumption it is true, and only continuity on $f:[0,\infty)\rightarrow [0,\infty)$ is needed. Of course, it is sufficient to show that some subsequence of $f(u_k)$ converges. So we can also assume w.l.o.g. that $u_k$ converges a.e. to $u$. 
Consider the sequence of non-negative measurable functions on $K$,  $v_k:=f\circ u_k$. Because $f$ is continuous, it converges a.e. to $v:=f\circ u$, and by the extra hypothesis, $\int_ K v_k \to\int_K v < \infty $. And, as an immediate consequence of the Fatou's lemma, this also implies $\int_ S v_k \to\int_S v$ for every measurable $S\subset K$ (this is the key point, after all, coming from the equality   $\int_K v_k =\int_S v_k +\int_{K\setminus S} v_k $: if one does not lose  mass globally, one does not lose mas locally). 
Now, as a general fact, on a finite measure space $K$ this situation 
implies the $L^1$ convergence. 
Indeed, let be given a number $\epsilon  > 0$. There is a number $\delta > 0 $ such that $\int_ S v < \epsilon$ whenever $S\subset K$ has measure $\big| S\big| < \delta$.
By the Severini-Egorov theorem $v_k$ converges almost uniformly, so there is some $S\subset K$ of measure less than $\delta$ such that $v_k$ converges uniformly to $v$ on $K\setminus S$. So we have:
$$  \|v _ k - v\| _ {1,K}=\|v  _k-v\|  _{1,S}+\|v  _k-v\|  _{1,K\setminus S}$$ $$\le \int _S v _ k +  \int _S v + \big|K\big| \|v  _ k-v\| _ {\infty,K\setminus S}  \le 2\epsilon + o(1)\, , \quad  \mathrm{as}\quad k\to\infty  \, .$$
Since this is true for any $\epsilon > 0$ we have   $\limsup_{k\to\infty}  \|v _ k - v\| _ {1,K}=0$ proving the convergence . Actually, the finiteness assumption on $K$ may also be dropped.
Here's  another proof: consider the sequence $ w_k : =  \sqrt v_k \in L^2(K)$. It is norm-bounded, converges to $w : = \sqrt{v}$ a.e., hence weakly in $L^2$; moreover, 
$\|w_ k \| _ 2 \to \|w \| _ 2 $ . In a Hilbert space, this implies strong convergence (this is immediately seen just expanding $\|w _ k - w\| _ 2 ^2$). The map $L^2\ni w\mapsto w^2\in L^1$ is continuous, and we conclude  $v_k \to v$ in $L^1$.
